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The problem is solve/simplify the following expression:

$2•2^\frac{202021^{202021} + 1}{2} \mod 202021$

I start by noticing that 202021 is prime. Then by Euler’s totient theorem:

$a^{\phi(n)} \cong 1 \mod n$

So, $2^{202020} \cong 1 \mod 202021$

However I’m unsure of where to go from here. I’d appreciate any help!

Bill Dubuque
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冥王 Hades
  • 3,083
  • Applying mod order reduction as explained in the dupe, we can reduce the exponent to $,{\large \frac{\color{#c00}p^p+1}2}\bmod{\color{#0a0}{p!-!1}},,$ whose calculation is trivial: hint $,\color{#c00}p\equiv 1\pmod{!\color{#0a0}{p!-!1}}\ \ $ – Bill Dubuque Mar 12 '24 at 22:35
  • @BillDubuque I see. For the exponent, after applying mod order reduction, I get $2^{101012}$, the resulting expression is $2^{101012} \mod 202021$ which simplifies to $202017$ for me. Can you confirm if this is correct? – 冥王 Hades Mar 12 '24 at 23:03
  • Yes, that's correct. Which way did you compute the exponent $\bmod 202020?$ It's not completely trivial due to the division by $2.\ $ Did you use Euler/QR to finish? – Bill Dubuque Mar 13 '24 at 00:00
  • @BillDubuque Yes. I noticed that $202021$ can be reduced modulo $202020$ using your hint as $202021 \equiv 1 \mod 202020$, hence $202021^{202021} \equiv 1^{202021} \mod 202020$ which is just $1$. The exponent is reduced to merely $\frac{1+3}{2}$. Was I correct? – 冥王 Hades Mar 13 '24 at 00:13
  • No, we need to double the modulus to balance the division by $2$, see here. $\ \ $ – Bill Dubuque Mar 13 '24 at 01:53
  • That would leave $\mod 404040$, but I’m not sure what follows, since $202021^{202021} \mod 404040$ would still be $1$. @BillDubuque – 冥王 Hades Mar 13 '24 at 02:21
  • No, $,n := p^p\equiv 1\pmod{!p!-!1},$ so $,n\equiv 1,$ or $,\color{#c00}p\pmod{!2(p!-!1)}.,$ It must be $,n\equiv\color{#c00}p,$ since only that is correct $!\bmod 8,$ (where $,p^2\equiv 1\Rightarrow p^{\rm odd}\equiv p).,$ Thus $,k := p^p+1\equiv \color{#c00}p+1\pmod{2(p!-!1))},,$ so $,k/2\equiv (p+1)/2\pmod{!p!-!1}\ \ $ – Bill Dubuque Mar 14 '24 at 06:48
  • Or by $\rm \mu LTE!:$ $\bmod p!-!1!:\ \color{#0a0}{p\equiv 1},$ so $\bmod{\color{#c00}2(p!-!1)}!:\ \color{#0a0}{p^{\large\color{#c00}2}\equiv 1^{\large\color{#c00}2}},,$ so $,p^{\large p}\equiv p^{\large 1+2k}\equiv p(\color{#0a0}{p^{\large\color{#c00}2}})^k\equiv p\ \ $ – Bill Dubuque Mar 14 '24 at 18:50

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